I have a template sequence like this:
val template = Seq(0,0,0,0,0,0,0,0,0,0)
I have another sequence which contains the index should be modified, like this:
val indices = Seq(1,3,7)
I want to modify the template sequence, in the way that, if the element's index is in the indices sequence, then replace the element to 1.
So the output should look like:
(0,1,0,1,0,0,0,1,0,0)
What is the simplest way to implement this function?
Figured it by myself:
indices.foldLeft(template)((b, a) => b.updated(a, 1))
The foldLeft in your solution repeatedly rebuilds the sequence. This could be inefficient if the sequence is for example a List and all indices point to the end of the list.
You could also achieve it without repeatedly rebuilding the collection:
val indexSet = indices.toSet
val result = template.zipWithIndex.map{ case (v, i) =>
if (indexSet contains i) 1 else v
}
For a List this solution would be amortized O(n), because checking whether an integer is in the set of indices is amortized O(1).
Maybe there are also use-cases where your solution is preferable, it's not so clear for general Seqs.
Related
I have a input file like this:
The Works of Shakespeare, by William Shakespeare
Language: English
and I want to use flatMap with the combinations method to get the K-V pairs per line.
This is what I do:
var pairs = input.flatMap{line =>
line.split("[\\s*$&#/\"'\\,.:;?!\\[\\(){}<>~\\-_]+")
.filter(_.matches("[A-Za-z]+"))
.combinations(2)
.toSeq
.map{ case array => array(0) -> array(1)}
}
I got 17 pairs after this, but missed 2 of them: (by,shakespeare) and (william,shakespeare). I think there might be something wrong with the last word of the first sentence, but I don't know how to solve it, can anyone tell me?
The combinations method will not give duplicates even if the values are in the opposite order. So the values you are missing already appear in the solution in the other order.
This code will create all ordered pairs of words in the text.
for {
line <- input
t <- line.split("""\W+""").tails if t.length > 1
a = t.head
b <- t.tail
} yield a -> b
Here is the description of the tails method:
Iterates over the tails of this traversable collection. The first value will be this traversable collection and the final one will be an empty traversable collection, with the intervening values the results of successive applications of tail.
I have a List[Either[Error, Satisfaction]], and I need to take the first element in that list that isRight. If there are no elements that satisfy isRight, then I should return the last Error.
For now I've thought on a takeWhile(_.isLeft), find the size of this last list, and then get the nth element in the original list, with n being the size of the first-lefts list, but I think there's a better solution for this.
Any help will be strongly appreciated.
You might reduce the list like so:
lst.reduceLeft((a,b) => if (a.isRight) a else b)
The result will be the first Right element or, if there is none, the final Left element.
You can do this, which has the dual problem to #jwvh's answer: if rest is empty, it has to go over the list again.
val (lefts, rest) = list.span(_.isLeft)
rest.headOption.getOrElse(lefts.last)
// or orElse(lefts.lastOption) if list can be empty
And the recursive solution without either problem for completeness (assuming non-empty list):
def foo(list: List[Either[Error, Satisfaction]]) = list match {
case x :: tail => if (x.isRight || tail.isEmpty) x else foo(tail)
}
Not that much more code.
In Scala, how to efficiently compare the contents of two lists/seqs, regardless of their order, without sorting (I don't know what the type of elements is)?
The lists/seqs may contain duplicates.
I have seen a somewhat similar discussion, but some answers there are incorrect, or they require sorting.
You can do
list1.groupBy(identity) == list2.groupBy(identity)
It's O(n).
If creating the temporary lists is an issue for you could create a helper method to get only the count for each item and not all occurrences:
def counter[T](l: List[T]) =
l.foldLeft(Map[T,Int]() withDefaultValue 0){ (m,x) =>
m + (x -> (1 + m(x)))
}
counter(list1) == counter(list2)
I'm new to functional programming, so some problems seems harder to solve using functional approach.
Let's say I have a list of numbers, like 1 to 10.000, and I want to get the items of the list which sums up to at most a number n (let's say 100). So, it would get the numbers until their sum is greater than 100.
In imperative programming, it's trivial to solve this problem, because I can keep a variable in each interaction, and stop once the objective is met.
But how can I do the same in functional programming? Since the sum function operates on completed lists, and I still don't have the completed list, how can I 'carry on' the computation?
If sum was lazily computed, I could write something like that:
(1 to 10000).sum.takeWhile(_ < 100)
P.S.:Even though any answer will be appreciated, I'd like one that doesn't compute the sum each time, since obviously the imperative version will be much more optimal regarding speed.
Edit:
I know that I can "convert" the imperative loop approach to a functional recursive function. I'm more interested in finding if one of the existing library functions can provide a way for me not to write one each time I need something.
Use Stream.
scala> val ss = Stream.from(1).take(10000)
ss: scala.collection.immutable.Stream[Int] = Stream(1, ?)
scala> ss.scanLeft(0)(_ + _)
res60: scala.collection.immutable.Stream[Int] = Stream(0, ?)
scala> res60.takeWhile(_ < 100).last
res61: Int = 91
EDIT:
Obtaining components is not very tricky either. This is how you can do it:
scala> ss.scanLeft((0, Vector.empty[Int])) { case ((sum, compo), cur) => (sum + cur, compo :+ cur) }
res62: scala.collection.immutable.Stream[(Int, scala.collection.immutable.Vector[Int])] = Stream((0,Vector()), ?)
scala> res62.takeWhile(_._1 < 100).last
res63: (Int, scala.collection.immutable.Vector[Int]) = (91,Vector(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13))
The second part of the tuple is your desired result.
As should be obvious, in this case, building a vector is wasteful. Instead we can only store the last number from the stream that contributed to sum.
scala> ss.scanLeft(0)(_ + _).zipWithIndex
res64: scala.collection.immutable.Stream[(Int, Int)] = Stream((0,0), ?)
scala> res64.takeWhile(_._1 < 100).last._2
res65: Int = 13
The way I would do this is with recursion. On each call, add the next number. Your base case is when the sum is greater than 100, at which point you return all the way up the stack. You'll need a helper function to do the actual recursion, but that's no big deal.
This isn't hard using "functional" methods either.
Using recursion, rather than maintaining your state in a local variable that you mutate, you keep it in parameters and return values.
So, to return the longest initial part of a list whose sum is at most N:
If the list is empty, you're done; return the empty list.
If the head of the list is greater than N, you're done; return the empty list.
Otherwise, let H be the head of the list.
All we need now is the initial part of the tail of the list whose sum is at most N - H, then we can "cons" H onto that list, and we're done.
We can compute this recursively using the same procedure as we have used this far, so it's an easy step.
A simple pseudocode solution:
sum_to (n, ls) = if isEmpty ls or n < (head ls)
then Nil
else (head ls) :: sum_to (n - head ls, tail ls)
sum_to(100, some_list)
All sequence operations which require only one pass through the sequence can be implemented using folds our reduce like it is sometimes called.
I find myself using folds very often since I became used to functional programming
so here odd one possible approach
Use an empty collection as initial value and fold according to this strategy
Given the processed collection and the new value check if their sum is low enough and if then spend the value to the collection else do nothing
that solution is not very efficient but I want to emphasize the following
map fold filter zip etc are the way to get accustomed to functional programming try to use them as much as possible instead of loping constructs or recursive functions your code will be more declarative and functional
Say I have a function, for example the old favourite
def factorial(n:Int) = (BigInt(1) /: (1 to n)) (_*_)
Now I want to find the biggest value of n for which factorial(n) fits in a Long. I could do
(1 to 100) takeWhile (factorial(_) <= Long.MaxValue) last
This works, but the 100 is an arbitrary large number; what I really want on the left hand side is an infinite stream that keeps generating higher numbers until the takeWhile condition is met.
I've come up with
val s = Stream.continually(1).zipWithIndex.map(p => p._1 + p._2)
but is there a better way?
(I'm also aware I could get a solution recursively but that's not what I'm looking for.)
Stream.from(1)
creates a stream starting from 1 and incrementing by 1. It's all in the API docs.
A Solution Using Iterators
You can also use an Iterator instead of a Stream. The Stream keeps references of all computed values. So if you plan to visit each value only once, an iterator is a more efficient approach. The downside of the iterator is its mutability, though.
There are some nice convenience methods for creating Iterators defined on its companion object.
Edit
Unfortunately there's no short (library supported) way I know of to achieve something like
Stream.from(1) takeWhile (factorial(_) <= Long.MaxValue) last
The approach I take to advance an Iterator for a certain number of elements is drop(n: Int) or dropWhile:
Iterator.from(1).dropWhile( factorial(_) <= Long.MaxValue).next - 1
The - 1 works for this special purpose but is not a general solution. But it should be no problem to implement a last method on an Iterator using pimp my library. The problem is taking the last element of an infinite Iterator could be problematic. So it should be implemented as method like lastWith integrating the takeWhile.
An ugly workaround can be done using sliding, which is implemented for Iterator:
scala> Iterator.from(1).sliding(2).dropWhile(_.tail.head < 10).next.head
res12: Int = 9
as #ziggystar pointed out, Streams keeps the list of previously computed values in memory, so using Iterator is a great improvment.
to further improve the answer, I would argue that "infinite streams", are usually computed (or can be computed) based on pre-computed values. if this is the case (and in your factorial stream it definately is), I would suggest using Iterator.iterate instead.
would look roughly like this:
scala> val it = Iterator.iterate((1,BigInt(1))){case (i,f) => (i+1,f*(i+1))}
it: Iterator[(Int, scala.math.BigInt)] = non-empty iterator
then, you could do something like:
scala> it.find(_._2 >= Long.MaxValue).map(_._1).get - 1
res0: Int = 22
or use #ziggystar sliding solution...
another easy example that comes to mind, would be fibonacci numbers:
scala> val it = Iterator.iterate((1,1)){case (a,b) => (b,a+b)}.map(_._1)
it: Iterator[Int] = non-empty iterator
in these cases, your'e not computing your new element from scratch every time, but rather do an O(1) work for every new element, which would improve your running time even more.
The original "factorial" function is not optimal, since factorials are computed from scratch every time. The simplest/immutable implementation using memoization is like this:
val f : Stream[BigInt] = 1 #:: (Stream.from(1) zip f).map { case (x,y) => x * y }
And now, the answer can be computed like this:
println( "count: " + (f takeWhile (_<Long.MaxValue)).length )
The following variant does not test the current, but the next integer, in order to find and return the last valid number:
Iterator.from(1).find(i => factorial(i+1) > Long.MaxValue).get
Using .get here is acceptable, since find on an infinite sequence will never return None.